# How small can a set admitting a nonatomic finite measure be?

Is it consistent that there exists a nonzero atomless finite measure on some $\sigma$-algebra on a cardinal $\kappa$ satisfying $\kappa<\mathfrak{c}$? Can there be such a measure on $\omega_1$ when $\omega_1<\mathfrak{c}$?

I would generally like to know where one can find related results.

My motivation comes from people in mathematical economics using the term "continuum of agents" for models in which agents are modeled by nonatomic probability spaces. I'd like to know if there couldn't be fewer agents than a continuum with this formulation.

• Sure, the continuum can be above strictly above the least real-valued measurable. – Asaf Karagila Jun 18 '18 at 7:27
• @AsafKaragila Thanks, good point! I was only thinking about smaller $\sigma$-algebras. – Michael Greinecker Jun 18 '18 at 7:30
• If you add lots of random reals, then the ground model $\mathbb R$ still has full measure. Then consider the Borel sets intersected with $\mathbb R^V$. Does this work? – Monroe Eskew Jun 18 '18 at 9:32

## 1 Answer

If $X \subseteq [0, 1]$ is a non Lebesgue null set of reals, then $B \cap X \mapsto \mu(B)$ (where $B$ is Borel and $\mu$ is Lebesgue measure) is an atomless measure on $X$. Conversely, if $m$ is an atomless probability measure on $\kappa$, then there is an inverse measure preserving map $f:\kappa \to [0, 1]$ - For every Borel $B$, $\mu(B) = m(f^{-1}[B])$. It follows that the least $\kappa$ for which there is an atomless probability measure on some sigma-algebra on $\kappa$ is the least cardinality of a non (Lebesgue) null set of reals which (as Monroe Eskew pointed out in the comments) is $\omega_1$ in the random real model.